The sum of the roots of the quadratic P(x)=ax^2+bx+c is -b/a
If we expand the sides of the inequality as subtract the terms from the right we get
ax^6+ax^4+bx^3+(-a-b)x^2+bx≥0
x(ax^5+ax^3+bx^2+(-a-b)x+b)≥0
For an inequality of the form pq≥0 to hold it must be true that if p>0, q>0 and if p<0, q<0 and if p=0, q=0. In other words
ax^5+ax^3+bx^2+(-a-b)x+b must equal 0 if x=0
this means b=0, so the sum of the roots of P = -b/a=0
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Posted by Jer
on 2021-04-02 10:18:00 |
Solution
